In a parallel plate air capacitor, the distance between plates is reduced to one-fourth and the space between them is filled with a dielectric medium of constant $2$. If the initial capacity of the capacitor is $4 \mu F$, then its new capacity is: (in $\mu F$)

$A$ parallel plate capacitor is connected to a battery. If a dielectric slab is inserted between the plates,which of the following quantities increases?

$A$ parallel plate capacitor with air as dielectric has a capacitance of $4 \mu F$. The space between the plates of the capacitor is completely filled with a material of dielectric constant $K = 5$ and charged to a potential of $100 \ V$. The work done to completely remove the dielectric material after the capacitor is disconnected from the battery is (in $J$)

The capacitance of a parallel plate capacitor is $2.5 \mu F$. When it is half filled with a dielectric as shown in the figure,its capacitance becomes $5 \mu F$. The dielectric constant of the dielectric is

Write the formula for the capacitance of a capacitor having a dielectric constant $K = 2$.

$A$ parallel plate capacitor with plate area $A$ and separation $d$ is filled with two dielectric materials as shown in the figure. The dielectric constants are $K_1$ and $K_2$ respectively. The capacitance will be: